solution1 - STA 6326 CHAPTER 1 - SOLUTIONS Problem 1.3 (c)...

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STA 6326 CHAPTER 1 - SOLUTIONS Problem 1.3 (c) Formally, ( 29 { } : c A B x S x A B = and { } : and c c A B x S x A x B = . Let ( 29 c x A B . Since x A B then x A and x B . Therefore, c x A and c x B . Hence, ( 29 c c x A B . Consequently, ( 29 ( 29 c c c A B A B . Let ( 29 c c x A B . Since x A and x B then ( 29 x A B . Therefore, ( 29 c x A B . Consequently, ( 29 ( 29 c c c A B A B . Since ( 29 ( 29 c c c A B A B and ( 29 ( 29 c c c A B A B then ( 29 ( 29 c c c A B A B = . Problem 1.5 (a) A B C = {identical twin females} (b) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 1 1 1 90 3 2 540 P A B C P C P B C P A B C P C P B C P A B = = = × × = . Problem 1.8 (a) Note that P (scoring i points) = Area of region Area of dart board i = Area of circle radius (6 ) - Area of circle radius (6 ( 1)) 5 5 Area of dart board r r i i - - + © 2009 David M. Nickerson, UCF

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= 2 2 2 )) 1 ( 6 ( 5 ) 6 ( 5 r i r i r π + - π - - π = ) 2 11 ( 25 1 i - (b) Since P (scoring i points) = ) 2 11 ( 25 1 i - then d di P (scoring i points) = 0 25 2 < - Hence, P (scoring i points) is a decreasing function of i . Problem 1.10 Let n = 2. Then by Theorem 1.1.4 ( 29 K n i c i c c c c n i i A A A A A A 1 2 1 2 1 1 = = = = = Now assume that for arbitrary n , K n i c i c n i i A A 1 1 = = = Then K K K 1 1 1 1 1 1 1 1 1 1 + = + = + = + = + = = = = = n i c i c n n i c i c n c n i i c n n i i c n i i A A A A A A A A Hence, by induction, K n i c i c n i i A A 1 1 = = = © 2009 David M. Nickerson, UCF 2
is true for all n . Problem 1.14 Suppose { } 1 2 , , , n S a a a = K . Then # of subsets of S = [ ] 0 0 # of subsets of size i n n i i n i = =   =     However, by the binomial theorem, ( 29 0 n n i n i i n a b a b i - =   + =     Therefore, 0 0 (1) (1) (1 1) 2 n n i n i n n i i n n i i - = =     = = + =         Problem 1.17 Since the pair is not ordered and the same number can appear twice, we have # of pieces = # of ways of selecting 2 items from n with replacement 2 1 2 n + - = 1 2 n + = ( 1) 2 n n + = . Problem 1.18 Define © 2009 David M. Nickerson, UCF 3

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S = {collection of all possible arrangements of n distinguishable balls in n cells} and A = {collection of arrangements in which exactly 1 cell is empty} Then to determine the number of elements in S imagine that each of the n distinguishable balls could be assigned n ways, i.e., to 1 of n different cells. Consequently, by the Fundamental Theorem of Counting we have n S n n n
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This note was uploaded on 10/03/2011 for the course STA 6246 taught by Professor Staff during the Spring '08 term at University of Central Florida.

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solution1 - STA 6326 CHAPTER 1 - SOLUTIONS Problem 1.3 (c)...

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